Question: Divide the following complex numbers. $ \dfrac{-15+3i}{-3-3i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-3+3i}$ $ \dfrac{-15+3i}{-3-3i} = \dfrac{-15+3i}{-3-3i} \cdot \dfrac{{-3+3i}}{{-3+3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-15+3i) \cdot (-3+3i)} {(-3-3i) \cdot (-3+3i)} = \dfrac{(-15+3i) \cdot (-3+3i)} {(-3)^2 - (-3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-15+3i) \cdot (-3+3i)} {(-3)^2 - (-3i)^2} = $ $ \dfrac{(-15+3i) \cdot (-3+3i)} {9 + 9} = $ $ \dfrac{(-15+3i) \cdot (-3+3i)} {18} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-15+3i}) \cdot ({-3+3i})} {18} = $ $ \dfrac{{-15} \cdot {(-3)} + {3} \cdot {(-3) i} + {-15} \cdot {3 i} + {3} \cdot {3 i^2}} {18} $ Evaluate each product of two numbers. $ \dfrac{45 - 9i - 45i + 9 i^2} {18} $ Finally, simplify the fraction. $ \dfrac{45 - 9i - 45i - 9} {18} = \dfrac{36 - 54i} {18} = 2-3i $